3^2x(2^x)=1/18

Solution for 3^2x(2^x)=1/18 equation:

3^2x(2^x)=1/18

We move all terms to the left:

3^2x(2^x)-(1/18)=0

We add all the numbers together, and all the variables

3^2x2^x-(+1/18)=0

We get rid of parentheses

3^2x2^x-1/18=0

We multiply all the terms by the denominator


3^2x2^x*18-1=0

Wy multiply elements

54x^2-1=0

a = 54; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·54·(-1)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*54}=\frac{0-6\sqrt{6}}{108}
=-\frac{6\sqrt{6}}{108}
=-\frac{\sqrt{6}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*54}=\frac{0+6\sqrt{6}}{108}
=\frac{6\sqrt{6}}{108}
=\frac{\sqrt{6}}{18}
$